$P[X \geq n] = \int_{X \geq n} dP$, given that $E[X]$ exists, and $X$ is a nonnegative random variable: why is this always true?

Asked Nov 18 '23 at 16:28

Active Nov 18 '23 at 16:28

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In this post, there's an equality:

$P[X \geq n] = \int_{X \geq n} dP$, given that $E[X]$ exists, and $X$ is a nonnegative random variable.

Could someone please explain why the equality holds? Why are we allowed to put the (arbitrary) probability function $P$ under the differential? I have looked at the definitions of probability and distribution function, but they didn't help to see why the equality is true.

Thank you.

asked Nov 18 '23 at 16:28

fragileradius

1,207

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This is actually how the Lebesgue integral is constructed (i.e $\int 1_A dP = P(A)$ for any measurable set $A$) – Giorgos Giapitzakis Nov 18 '23 at 16:36

$P[X \geq n] = \int_{X \geq n} dP$, given that $E[X]$ exists, and $X$ is a nonnegative random variable: why is this always true?

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